Word Problems of fractions worksheets - Math Worksheets ... - Free Printable

Let's solve each problem step by step.

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Problem 1: Total yield of pumpkins

Question: The big field yields \(3 \frac{2}{5}\) tons of pumpkins, and the small field yields \(2 \frac{1}{12}\) tons of pumpkins. What is the total yield of pumpkins?

#### Solution:
1. Convert mixed numbers to improper fractions:
\[
3 \frac{2}{5} = \frac{3 \times 5 + 2}{5} = \frac{15 + 2}{5} = \frac{17}{5}
\]
\[
2 \frac{1}{12} = \frac{2 \times 12 + 1}{12} = \frac{24 + 1}{12} = \frac{25}{12}
\]

2. Add the two fractions:
\[
\frac{17}{5} + \frac{25}{12}
\]
Find the least common denominator (LCD) of 5 and 12, which is 60.

3. Rewrite each fraction with the LCD:
\[
\frac{17}{5} = \frac{17 \times 12}{5 \times 12} = \frac{204}{60}
\]
\[
\frac{25}{12} = \frac{25 \times 5}{12 \times 5} = \frac{125}{60}
\]

4. Add the fractions:
\[
\frac{204}{60} + \frac{125}{60} = \frac{204 + 125}{60} = \frac{329}{60}
\]

5. Convert the improper fraction back to a mixed number:
\[
\frac{329}{60} = 5 \frac{29}{60}
\]

Answer:
\[
\boxed{5 \frac{29}{60}}
\]

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Problem 2: Average weight of zucchinis

Question: The biggest zucchini weighs \(2 \frac{5}{8}\) pounds, which is \(1 \frac{1}{12}\) pounds more than the average weight of zucchinis. What is the average weight of zucchinis?

#### Solution:
1. Convert mixed numbers to improper fractions:
\[
2 \frac{5}{8} = \frac{2 \times 8 + 5}{8} = \frac{16 + 5}{8} = \frac{21}{8}
\]
\[
1 \frac{1}{12} = \frac{1 \times 12 + 1}{12} = \frac{12 + 1}{12} = \frac{13}{12}
\]

2. Let the average weight of zucchinis be \(x\). According to the problem:
\[
x + \frac{13}{12} = \frac{21}{8}
\]

3. Solve for \(x\):
\[
x = \frac{21}{8} - \frac{13}{12}
\]

4. Find the least common denominator (LCD) of 8 and 12, which is 24.

5. Rewrite each fraction with the LCD:
\[
\frac{21}{8} = \frac{21 \times 3}{8 \times 3} = \frac{63}{24}
\]
\[
\frac{13}{12} = \frac{13 \times 2}{12 \times 2} = \frac{26}{24}
\]

6. Subtract the fractions:
\[
x = \frac{63}{24} - \frac{26}{24} = \frac{63 - 26}{24} = \frac{37}{24}
\]

7. Convert the improper fraction back to a mixed number:
\[
\frac{37}{24} = 1 \frac{13}{24}
\]

Answer:
\[
\boxed{1 \frac{13}{24}}
\]

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Problem 3: Soil used in this month

Question: Farm Joe ordered 3 bags of soil, each weighing \(4 \frac{2}{5}\) kilograms. He used the first bag in a week. At the end of the month, there were \(2 \frac{3}{4}\) kilograms left in the second bag and \(\frac{7}{8}\) kilograms left in the third bag. How much soil was used in this month?

#### Solution:
1. Convert mixed numbers to improper fractions:
\[
4 \frac{2}{5} = \frac{4 \times 5 + 2}{5} = \frac{20 + 2}{5} = \frac{22}{5}
\]
\[
2 \frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{8 + 3}{4} = \frac{11}{4}
\]

2. Calculate the total amount of soil ordered:
\[
3 \times \frac{22}{5} = \frac{66}{5}
\]

3. Calculate the amount of soil left in the second and third bags:
\[
\text{Soil left in the second bag} = \frac{11}{4}
\]
\[
\text{Soil left in the third bag} = \frac{7}{8}
\]

4. Find the least common denominator (LCD) of 4 and 8, which is 8. Rewrite \(\frac{11}{4}\) with the LCD:
\[
\frac{11}{4} = \frac{11 \times 2}{4 \times 2} = \frac{22}{8}
\]

5. Add the soil left in the second and third bags:
\[
\frac{22}{8} + \frac{7}{8} = \frac{22 + 7}{8} = \frac{29}{8}
\]

6. Calculate the total soil used:
\[
\text{Total soil used} = \text{Total soil ordered} - \text{Soil left}
\]
\[
\text{Total soil used} = \frac{66}{5} - \frac{29}{8}
\]

7. Find the least common denominator (LCD) of 5 and 8, which is 40. Rewrite both fractions with the LCD:
\[
\frac{66}{5} = \frac{66 \times 8}{5 \times 8} = \frac{528}{40}
\]
\[
\frac{29}{8} = \frac{29 \times 5}{8 \times 5} = \frac{145}{40}
\]

8. Subtract the fractions:
\[
\frac{528}{40} - \frac{145}{40} = \frac{528 - 145}{40} = \frac{383}{40}
\]

9. Convert the improper fraction back to a mixed number:
\[
\frac{383}{40} = 9 \frac{23}{40}
\]

Answer:
\[
\boxed{9 \frac{23}{40}}
\]

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Problem 4: Price of carrots this month

Question: Last month, the price of one pound of carrots was \$\(2 \frac{1}{5}\), and Joe sold \(12 \frac{1}{12}\) pounds. This month, the price increased by \$\(1 \frac{1}{10}\), and Farmer Joe only sold \(5 \frac{1}{8}\) pounds. What is the price of a pound of carrots this month?

#### Solution:
1. Convert mixed numbers to improper fractions:
\[
2 \frac{1}{5} = \frac{2 \times 5 + 1}{5} = \frac{10 + 1}{5} = \frac{11}{5}
\]
\[
1 \frac{1}{10} = \frac{1 \times 10 + 1}{10} = \frac{10 + 1}{10} = \frac{11}{10}
\]

2. Calculate the new price of carrots this month:
\[
\text{New price} = \text{Old price} + \text{Increase}
\]
\[
\text{New price} = \frac{11}{5} + \frac{11}{10}
\]

3. Find the least common denominator (LCD) of 5 and 10, which is 10. Rewrite \(\frac{11}{5}\) with the LCD:
\[
\frac{11}{5} = \frac{11 \times 2}{5 \times 2} = \frac{22}{10}
\]

4. Add the fractions:
\[
\frac{22}{10} + \frac{11}{10} = \frac{22 + 11}{10} = \frac{33}{10}
\]

5. Convert the improper fraction back to a mixed number:
\[
\frac{33}{10} = 3 \frac{3}{10}
\]

Answer:
\[
\boxed{3 \frac{3}{10}}
\]

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Problem 5: Crates of tomatoes left

Question: There were \(24 \frac{1}{4}\) crates of tomatoes in the barn. \(7 \frac{3}{5}\) crates were rotten and had to be thrown out. Joe sold \(8 \frac{1}{3}\) crates and canned \(7 \frac{5}{6}\) crates. How many crates of tomatoes were left?

#### Solution:
1. Convert mixed numbers to improper fractions:
\[
24 \frac{1}{4} = \frac{24 \times 4 + 1}{4} = \frac{96 + 1}{4} = \frac{97}{4}
\]
\[
7 \frac{3}{5} = \frac{7 \times 5 + 3}{5} = \frac{35 + 3}{5} = \frac{38}{5}
\]
\[
8 \frac{1}{3} = \frac{8 \times 3 + 1}{3} = \frac{24 + 1}{3} = \frac{25}{3}
\]
\[
7 \frac{5}{6} = \frac{7 \times 6 + 5}{6} = \frac{42 + 5}{6} = \frac{47}{6}
\]

2. Calculate the total number of crates removed (rotten, sold, and canned):
\[
\text{Total removed} = \frac{38}{5} + \frac{25}{3} + \frac{47}{6}
\]

3. Find the least common denominator (LCD) of 5, 3, and 6, which is 30. Rewrite each fraction with the LCD:
\[
\frac{38}{5} = \frac{38 \times 6}{5 \times 6} = \frac{228}{30}
\]
\[
\frac{25}{3} = \frac{25 \times 10}{3 \times 10} = \frac{250}{30}
\]
\[
\frac{47}{6} = \frac{47 \times 5}{6 \times 5} = \frac{235}{30}
\]

4. Add the fractions:
\[
\frac{228}{30} + \frac{250}{30} + \frac{235}{30} = \frac{228 + 250 + 235}{30} = \frac{713}{30}
\]

5. Calculate the number of crates left:
\[
\text{Crates left} = \text{Initial crates} - \text{Total removed}
\]
\[
\text{Crates left} = \frac{97}{4} - \frac{713}{30}
\]

6. Find the least common denominator (LCD) of 4 and 30, which is 60. Rewrite each fraction with the LCD:
\[
\frac{97}{4} = \frac{97 \times 15}{4 \times 15} = \frac{1455}{60}
\]
\[
\frac{713}{30} = \frac{713 \times 2}{30 \times 2} = \frac{1426}{60}
\]

7. Subtract the fractions:
\[
\frac{1455}{60} - \frac{1426}{60} = \frac{1455 - 1426}{60} = \frac{29}{60}
\]

Answer:
\[
\boxed{\frac{29}{60}}
\]

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Problem 6: Total hours the farmer's market is open

Question: The farmer's market opens for \(2 \frac{1}{5}\) hours in the morning and \(3 \frac{2}{3}\) hours in the afternoon. How long is the farmer's market open in a day?

#### Solution:
1. Convert mixed numbers to improper fractions:
\[
2 \frac{1}{5} = \frac{2 \times 5 + 1}{5} = \frac{10 + 1}{5} = \frac{11}{5}
\]
\[
3 \frac{2}{3} = \frac{3 \times 3 + 2}{3} = \frac{9 + 2}{3} = \frac{11}{3}
\]

2. Add the two fractions:
\[
\frac{11}{5} + \frac{11}{3}
\]
Find the least common denominator (LCD) of 5 and 3, which is 15.

3. Rewrite each fraction with the LCD:
\[
\frac{11}{5} = \frac{11 \times 3}{5 \times 3} = \frac{33}{15}
\]
\[
\frac{11}{3} = \frac{11 \times 5}{3 \times 5} = \frac{55}{15}
\]

4. Add the fractions:
\[
\frac{33}{15} + \frac{55}{15} = \frac{33 + 55}{15} = \frac{88}{15}
\]

5. Convert the improper fraction back to a mixed number:
\[
\frac{88}{15} = 5 \frac{13}{15}
\]

Answer:
\[
\boxed{5 \frac{13}{15}}
\]

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Final Answers:

1. \(\boxed{5 \frac{29}{60}}\)
2. \(\boxed{1 \frac{13}{24}}\)
3. \(\boxed{9 \frac{23}{40}}\)
4. \(\boxed{3 \frac{3}{10}}\)
5. \(\boxed{\frac{29}{60}}\)
6. \(\boxed{5 \frac{13}{15}}\)